Distribution of combination of random variables

Question

I am struggling to find a solution on finding the following distribution: $P(X+Y \leq z) p + P(Y \leq z) (1-p)$ where $X, Y$ are both gaussian random variables with zero mean and variances $s_x$ and $s_Y$, respectively. $Y$ is statistically independent from $X$. Also, $p$ is a probability. Is this sum Gaussian also ? And if so, how can I find mean and variance? I've reached to the conclusion that X+Y is gaussian with zero mean and varriance $s_x + s_y$, therefore the whole sum is equal to a Gaussian with zero mean and varriance $p^2 (s_x + s_y)+(1-p)^2 s_y $. Is my thought correct?

Answer 1

What you have created is a weighted mixture of two Gaussian distributions with the same means of $0$ but different variances of $s_1+s_2$ and $s_2$. You should not expect this to have a Gaussian distribution; it does not.

If they had had different means of $m_1+m_2$ and $m_2$ then your distribution would have mean of $m = p(m_1+m_2)+(1-p)m_2 = pm_1+m_2$ and variance of $s=p(s_1+s_2)+p(m_1+m_2)^2 + (1-p)s_2+(1-p)m_2^2 -m^2 = ps_1+s_2+p(1-p)m_1^2$.

But since $m_1+m_2=m_2=0$, your distribution has mean $m=0$ and variance $s=ps_1+s_2$.